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‎doc/Programs/BlackScholes/bs.py‎

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import math
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from scipy.stats import norm
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def black_scholes_call_price(S, K, T, r, sigma):
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"""
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Analytical solution to the Black-Scholes formula for a European call option.
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Parameters:
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S - Current stock price
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K - Strike price
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T - Time to maturity (in years)
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r - Risk-free interest rate
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sigma - Volatility of the underlying asset
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Returns:
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Call option price
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"""
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d1 = (math.log(S / K) + (r + 0.5 * sigma ** 2) * T) / (sigma * math.sqrt(T))
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d2 = d1 - sigma * math.sqrt(T)
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call_price = S * norm.cdf(d1) - K * math.exp(-r * T) * norm.cdf(d2)
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return call_price
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# Example usage
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S = 100 # Stock price
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K = 100 # Strike price
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T = 1 # Time to maturity (1 year)
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r = 0.05 # Risk-free interest rate
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sigma = 0.2 # Volatility
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price = black_scholes_call_price(S, K, T, r, sigma)
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print(f"European Call Option Price: {price:.4f}")
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# simple numerical solution with plain explicit finite difference
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import numpy as np
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import matplotlib.pyplot as plt
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def black_scholes_fd_explicit(S_max, K, T, r, sigma, M=100, N=100):
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"""
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Finite Difference Method (Explicit Scheme) for Black-Scholes Equation
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Parameters:
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S_max - Maximum stock price considered
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K - Strike price
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T - Time to maturity
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r - Risk-free interest rate
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sigma - Volatility
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M - Number of time steps
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N - Number of price steps
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Returns:
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S - Stock prices
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V - Option values at t=0
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"""
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dt = T / M
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dS = S_max / N
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S = np.linspace(0, S_max, N + 1)
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V = np.maximum(S - K, 0) # Terminal payoff for a call option
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for j in range(M):
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V_old = V.copy()
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for i in range(1, N):
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delta = (V_old[i + 1] - V_old[i - 1]) / (2 * dS)
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gamma = (V_old[i + 1] - 2 * V_old[i] + V_old[i - 1]) / (dS ** 2)
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V[i] = V_old[i] + dt * (0.5 * sigma ** 2 * S[i] ** 2 * gamma +
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r * S[i] * delta - r * V_old[i])
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V[0] = 0 # Option worthless if stock price is 0
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V[-1] = S_max - K * np.exp(-r * (T - (j + 1) * dt)) # Approximate boundary condition
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return S, V
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# Parameters
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S_max = 200
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K = 100
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T = 1
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r = 0.05
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sigma = 0.2
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S, V = black_scholes_fd_explicit(S_max, K, T, r, sigma)
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plt.plot(S, V)
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plt.xlabel("Stock Price")
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plt.ylabel("Option Value")
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plt.title("European Call Option Price via Finite Difference Method")
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plt.grid(True)
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plt.show()

‎doc/Programs/BlackScholes/bsanalytic.py‎

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import math
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from scipy.stats import norm
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def black_scholes_call_price(S, K, T, r, sigma):
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"""
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Analytical solution to the Black-Scholes formula for a European call option.
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Parameters:
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S - Current stock price
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K - Strike price
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T - Time to maturity (in years)
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r - Risk-free interest rate
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sigma - Volatility of the underlying asset
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Returns:
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Call option price
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"""
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d1 = (math.log(S / K) + (r + 0.5 * sigma ** 2) * T) / (sigma * math.sqrt(T))
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d2 = d1 - sigma * math.sqrt(T)
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call_price = S * norm.cdf(d1) - K * math.exp(-r * T) * norm.cdf(d2)
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return call_price
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# Example usage
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S = 100 # Stock price
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K = 100 # Strike price
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T = 1 # Time to maturity (1 year)
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r = 0.05 # Risk-free interest rate
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sigma = 0.2 # Volatility
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price = black_scholes_call_price(S, K, T, r, sigma)
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print(f"European Call Option Price: {price:.4f}")
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# simple numerical solution with plain explicit finite difference
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import numpy as np
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import matplotlib.pyplot as plt
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def black_scholes_fd_explicit(S_max, K, T, r, sigma, M=100, N=100):
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"""
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Finite Difference Method (Explicit Scheme) for Black-Scholes Equation
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Parameters:
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S_max - Maximum stock price considered
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K - Strike price
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T - Time to maturity
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r - Risk-free interest rate
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sigma - Volatility
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M - Number of time steps
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N - Number of price steps
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Returns:
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S - Stock prices
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V - Option values at t=0
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"""
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dt = T / M
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dS = S_max / N
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S = np.linspace(0, S_max, N + 1)
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V = np.maximum(S - K, 0) # Terminal payoff for a call option
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for j in range(M):
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V_old = V.copy()
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for i in range(1, N):
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delta = (V_old[i + 1] - V_old[i - 1]) / (2 * dS)
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gamma = (V_old[i + 1] - 2 * V_old[i] + V_old[i - 1]) / (dS ** 2)
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V[i] = V_old[i] + dt * (0.5 * sigma ** 2 * S[i] ** 2 * gamma +
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r * S[i] * delta - r * V_old[i])
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V[0] = 0 # Option worthless if stock price is 0
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V[-1] = S_max - K * np.exp(-r * (T - (j + 1) * dt)) # Approximate boundary condition
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return S, V
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# Parameters
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S_max = 200
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K = 100
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T = 1
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r = 0.05
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sigma = 0.2
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S, V = black_scholes_fd_explicit(S_max, K, T, r, sigma)
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plt.plot(S, V)
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plt.xlabel("Stock Price")
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plt.ylabel("Option Value")
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plt.title("European Call Option Price via Finite Difference Method")
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plt.grid(True)
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plt.show()

‎doc/Programs/BlackScholes/bsfdm.py‎

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import math
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from scipy.stats import norm
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def black_scholes_call_price(S, K, T, r, sigma):
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"""
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Analytical solution to the Black-Scholes formula for a European call option.
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Parameters:
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S - Current stock price
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K - Strike price
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T - Time to maturity (in years)
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r - Risk-free interest rate
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sigma - Volatility of the underlying asset
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Returns:
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Call option price
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"""
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d1 = (math.log(S / K) + (r + 0.5 * sigma ** 2) * T) / (sigma * math.sqrt(T))
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d2 = d1 - sigma * math.sqrt(T)
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call_price = S * norm.cdf(d1) - K * math.exp(-r * T) * norm.cdf(d2)
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return call_price
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# Example usage
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S = 100 # Stock price
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K = 100 # Strike price
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T = 1 # Time to maturity (1 year)
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r = 0.05 # Risk-free interest rate
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sigma = 0.2 # Volatility
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price = black_scholes_call_price(S, K, T, r, sigma)
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print(f"European Call Option Price: {price:.4f}")
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# simple numerical solution with plain explicit finite difference
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import numpy as np
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import matplotlib.pyplot as plt
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def black_scholes_fd_explicit(S_max, K, T, r, sigma, M=100, N=100):
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"""
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Finite Difference Method (Explicit Scheme) for Black-Scholes Equation
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Parameters:
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S_max - Maximum stock price considered
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K - Strike price
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T - Time to maturity
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r - Risk-free interest rate
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sigma - Volatility
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M - Number of time steps
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N - Number of price steps
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Returns:
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S - Stock prices
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V - Option values at t=0
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"""
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dt = T / M
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dS = S_max / N
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S = np.linspace(0, S_max, N + 1)
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V = np.maximum(S - K, 0) # Terminal payoff for a call option
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for j in range(M):
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V_old = V.copy()
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for i in range(1, N):
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delta = (V_old[i + 1] - V_old[i - 1]) / (2 * dS)
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gamma = (V_old[i + 1] - 2 * V_old[i] + V_old[i - 1]) / (dS ** 2)
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V[i] = V_old[i] + dt * (0.5 * sigma ** 2 * S[i] ** 2 * gamma +
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r * S[i] * delta - r * V_old[i])
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V[0] = 0 # Option worthless if stock price is 0
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V[-1] = S_max - K * np.exp(-r * (T - (j + 1) * dt)) # Approximate boundary condition
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return S, V
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# Parameters
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S_max = 200
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K = 100
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T = 1
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r = 0.05
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sigma = 0.2
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S, V = black_scholes_fd_explicit(S_max, K, T, r, sigma)
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plt.plot(S, V)
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plt.xlabel("Stock Price")
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plt.ylabel("Option Value")
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plt.title("European Call Option Price via Finite Difference Method")
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plt.grid(True)
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plt.show()

‎doc/Programs/BlackScholes/bsmc.py‎

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# Monte Carlo implementation (With Antithetic Variates)
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import numpy as np
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def monte_carlo_black_scholes_option(S0, K, T, r, sigma,
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num_simulations=100000,
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option_type='call',
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use_antithetic=True):
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"""
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Monte Carlo simulation for European call/put option with variance reduction.
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Parameters:
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S0 - Initial stock price
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K - Strike price
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T - Time to maturity (in years)
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r - Risk-free interest rate
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sigma - Volatility
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num_simulations - Number of Monte Carlo simulations
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option_type - 'call' or 'put'
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use_antithetic - Whether to use antithetic variates for variance reduction
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Returns:
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Estimated option price
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"""
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n = num_simulations // 2 if use_antithetic else num_simulations
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Z = np.random.normal(size=n)
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# Simulate asset prices at maturity
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ST1 = S0 * np.exp((r - 0.5 * sigma**2) * T + sigma * np.sqrt(T) * Z)
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if use_antithetic:
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ST2 = S0 * np.exp((r - 0.5 * sigma**2) * T - sigma * np.sqrt(T) * Z)
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ST = np.concatenate([ST1, ST2])
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else:
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ST = ST1
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# Calculate payoff
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if option_type.lower() == 'call':
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payoff = np.maximum(ST - K, 0)
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elif option_type.lower() == 'put':
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payoff = np.maximum(K - ST, 0)
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else:
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raise ValueError("option_type must be 'call' or 'put'")
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# Discount to present value
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price = np.exp(-r * T) * np.mean(payoff)
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return price
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# Example usage
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S0 = 100
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K = 100
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T = 1
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r = 0.05
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sigma = 0.2
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call_price = monte_carlo_black_scholes_option(S0, K, T, r, sigma, option_type='call')
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put_price = monte_carlo_black_scholes_option(S0, K, T, r, sigma, option_type='put')
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print(f"Monte Carlo Estimated Call Price (Antithetic): {call_price:.4f}")
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print(f"Monte Carlo Estimated Put Price (Antithetic): {put_price:.4f}")

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